## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 424: 50

#### Answer

$$\frac{1}{2}\ln |\tan \theta-1|-\frac{1}{2}\ln |\tan \theta+1|+C$$

#### Work Step by Step

Given $$\int \frac{\sec ^{2} \theta d \theta}{\tan ^{2} \theta-1}$$ Let $$u=\tan \theta \ \ \ \ \ \to \ \ \ \ du=\sec ^{2} \theta d \theta$$ Then \begin{aligned} \int \frac{\sec ^{2} \theta d \theta}{\tan ^{2} \theta-1}&=\int \frac{ d u}{u^{2} -1} \end{aligned} Since \begin{align*} \frac{1}{(u^2-1)} &=\frac{A}{u-1}+\frac{B}{u+1}\\ &= \frac{A(u+1)+B(u-1)}{(u^2-1)}\\ 1&= A(u+1)+B(u-1) \end{align*} Then \begin{align*} \text{at } u&=1 \ \ \ \ \to A=1/2 \\ \text{at } u&=- 1\ \ \ \ \to B=-1/2 \end{align*} Hence \begin{aligned} \int \frac{\sec ^{2} \theta d \theta}{\tan ^{2} \theta-1}&=\int \frac{ d u}{u^{2} -1}\\ &= \frac{1}{2} \int \frac{du}{u-1}- \frac{1}{2} \int \frac{du}{u+1}\\ &=\frac{1}{2}\ln |u-1|-\frac{1}{2}\ln |u+1|+C\\ &=\frac{1}{2}\ln |\tan \theta-1|-\frac{1}{2}\ln |\tan \theta+1|+C \end{aligned}

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